9 another example; [made Community Wiki]

An example where $D=$3-ball and no $\mathbf{s}$ can existhelicity density must have a zero:

An example where $D=T^2\times (0,2\pi)$ and helicity density must have a zero:

Set $D=S^1\times S^1\times(0,2\pi)$ and let $(\theta,\zeta,r)$ be the obvious coordinate system. Set $B=f(r) dr\wedge d\theta+g(r) dr\wedge d\zeta$ where $$f(r)=\cos(2r),$$ and $$g(r)=\sin(r).$$ Clearly, $A=\frac{1}{2}\sin(2r)d\theta-\cos(r)d\zeta$ satisfies $B=dA$ and $B$ is nowhere vanishing. A quick calculation shows that $\int_D A\wedge B=0$.

Now suppose that $\mathbf{s}$ is an arbitrary closed 1-form. Either by using Stoke's theorem or by direct calculation, the fact that the total toroidal and poloidal fluxes, $2\pi\int_0^{2\pi}f(r)dr$ and $2\pi\int_0^{2\pi}g(r)dr$, are zero implies that $\int_D(A+\mathbf{s})\wedge B=0$. Thus, the helicity density must always have a zero.

8 slight generalization of question

Suppose you are given a nowhere-vanishing exact 2-form $B=dA$ on an open, connected domain $D\subset\mathbb{R}^3$. I'd like to think of $B$ as a magnetic field.

Consider the product $H(A)=A\wedge dA$. At least in the plasma physics literature, $H(A)$ is known as the magnetic helicity density.

How can one determine if there is a closed one-form $\mathbf{s}$ such that $H(A+\mathbf{s})$ is non-zero at all points in $D$?

The reason I am interested in this question is that if you can find such an $\mathbf{s}$, then $A+\mathbf{s}$ will define a contact structure on $D$ whose Reeb vector field gives the magnetic field lines. Thus, the question is closely related to the Hamiltonian structure of magnetic field line dynamics.

I'll elaborate on this last point a bit. If there is a vector potential $A$ such that $A∧dA$ is non-zero everywhere, then the distribution $ξ=ker(A)$ ξ=\text{ker}(A)$is nowhere integrable, meaning$ξ$defines a contact structure on$D$with a global contact 1-form$A$. The Reeb vector field of this contact structure relative to the contact form$A$is the unique vector field$X$that satisfies$A(X)=1$and$i_XdA=0$. Using the standard volume form$μ_o$,$dA$can be expressed as$i_B\mu_o$for a unique divergence-free vector field$\mathbf{B}$(I'm having trouble typing$\mathbf{B}$as a subscript). Thus, the second condition on the Reeb vector field can be expressed as$\mathbf{B}×X=0$, which implies the integral curves of X coincide with the magnetic field lines. More generally, suppose$M$is an orientable odd-dimensional manifold equipped with an exact 2-form$\omega$of maximal rank. Also assume that the characteristic line bundle associated to$\omega$admits a non-vanishing section$b:M\rightarrow \text{ker}(\omega)$. What is the obstruction to the existence of a 1-form$\vartheta$with$d\vartheta=\omega$and$\vartheta(b)>0$? some observations/comments: 1) If$A(\mathbf{B})$is bounded above and below on$D$, then a sufficient condition for there to be an$\mathbf{s}$that gives a nowhere-vanishing helicity density is the existence of a closed one-form$\alpha$with$\alpha(\mathbf{B})$nowhere vanishing. In that case,$\mathbf{s}=\lambda \alpha$, where$\lambda$is some large real number (with appropriate sign), would work. If there is such an$\alpha$, then, being closed, it defines a foliation whose leaves are transverse to the divergence-free field$\mathbf{B}$. I suspect the question that asks whether a given non-vanishing divergence-free vector field admits a transverse co-dimension one foliation has been studied before, but I am not familiar with any work of this type. An example where$D=$3-ball and no$\mathbf{s}$can exist: Let$D$consist of those points in$\mathbb{R}^3$with$x^2+y^2 < a^2$for a real number$a>1$. Note that all closed 1-forms are exact in this case. Let$f:[0,\infty)\rightarrow\mathbb{R}$be a smooth, non-decreasing function such that$f(r)=0$for$r<1/10$and$f(r)=1$for$r\ge1/2$. Let$g:\mathbb{R}\rightarrow \mathbb{R}$be the polynomial$g(r)=1-3r+2r^2$. Define the 2-form$B$using the divergence free vector field$\mathbf{B}(x,y,z)=f(\sqrt{x^2+y^2})e_\phi(x,y,z)+g(\sqrt{x^2+y^2})e_z$. Here$e_\phi$is the azimuthal unit vector and$e_z$is the$z$-directed unit vector. It is easy to verify that$B$, thus defined, is an exact 2-form that is nowhere vanishing. Because$g(1)=0$and$f(1)=1$, the circle,$C$, in the$z=0$-plane,$x^2+y^2=1$, is an integral curve for the vector field$\mathbf{B}$. I will use this fact to prove that the helicity density must have a zero for any choice of gauge. Let$A$satisfy$dA=B$and suppose$A\wedge B$is non-zero at all points in$D$. Note that$A\wedge B=A(\mathbf{B})\mu_o$, meaning$h=A(\mathbf{B})$is a nowhere vanishing function. Without loss of generality, I will assume$h>0$. Thus, the line integral$I=\oint_C h\frac{dl}{|\mathbf{B}|}$satisfies$I>0$. But, by Stoke's theorem,$I=2\pi\int_0^1g(r)rdr=0$, as is readily verified by directly evaluating the integral. Thus, there can be no such$A$. 7 corrected an error Suppose you are given a nowhere-vanishing exact 2-form$B=dA$on an open, connected domain$D\subset\mathbb{R}^3$. I'd like to think of$B$as a magnetic field. Consider the product$H(A)=A\wedge dA$. At least in the plasma physics literature,$H(A)$is known as the magnetic helicity density. How can one determine if there is a closed one-form$\mathbf{s}$such that$H(A+\mathbf{s})$is non-zero at all points in$D$? The reason I am interested in this question is that if you can find such an$\mathbf{s}$, then$A+\mathbf{s}$will define a contact structure on$D$whose Reeb vector field gives the magnetic field lines. Thus, the question is closely related to the Hamiltonian structure of magnetic field line dynamics. I'll elaborate on this last point a bit. If there is a vector potential$A$such that$A∧dA$is non-zero everywhere, then the distribution$ξ=ker(A)$is nowhere integrable, meaning$ξ$defines a contact structure on$D$with a global contact 1-form$A$. The Reeb vector field of this contact structure relative to the contact form$A$is the unique vector field$X$that satisfies$A(X)=1$and$i_XdA=0$. Using the standard volume form$μ_o$,$dA$can be expressed as$i_B\mu_o$for a unique divergence-free vector field$\mathbf{B}$(I'm having trouble typing$\mathbf{B}$as a subscript). Thus, the second condition on the Reeb vector field can be expressed as$\mathbf{B}×X=0$, which implies the integral curves of X coincide with the magnetic field lines. some observations/comments: 1) If$A(\mathbf{B})$is bounded above and below on$D$, then a sufficient condition for there to be an$\mathbf{s}$that gives a nowhere-vanishing helicity density is the existence of a closed one-form$\alpha$with$\alpha(\mathbf{B})$nowhere vanishing. In that case,$\mathbf{s}=\lambda \alpha$, where$\lambda$is some large real number (with appropriate sign), would work. If there is such an$\alpha$, then, being closed, it defines a foliation whose leaves are transverse to the divergence-free field$\mathbf{B}$. I suspect the question that asks whether a given non-vanishing divergence-free vector field admits a transverse co-dimension one foliation has been studied before, but I am not familiar with any work of this type. 2) If there is an invariant for the flow of$\mathbf{B}$,$\chi$, then one gauge-independent constraint on$A\wedge dA$is that the integral of the latter between any pair of compact regular level sets of$\chi$cannot depend on the choice of$\mathbf{s}$. In particular, if there are two compact regular level sets such that the integral of$A\wedge dA$between them vanishes, then there cannot be any gauge such that the magnetic helicity density is nowhere vanishing. An example where$D=$3-ball and no$\mathbf{s}$can exist: Let$D$consist of those points in$\mathbb{R}^3$with$x^2+y^2 < a^2$for a real number$a>1$. Note that all closed 1-forms are exact in this case. Let$f:[0,\infty)\rightarrow\mathbb{R}$be a smooth, non-decreasing function such that$f(r)=0$for$r<1/10$and$f(r)=1$for$r\ge1/2$. Let$g:\mathbb{R}\rightarrow \mathbb{R}$be the polynomial$g(r)=1-3r+2r^2$. Define the 2-form$B$using the divergence free vector field$\mathbf{B}(x,y,z)=f(\sqrt{x^2+y^2})e_\phi(x,y,z)+g(\sqrt{x^2+y^2})e_z$. Here$e_\phi$is the azimuthal unit vector and$e_z$is the$z$-directed unit vector. It is easy to verify that$B$, thus defined, is an exact 2-form that is nowhere vanishing. Because$g(1)=0$and$f(1)=1$, the circle,$C$, in the$z=0$-plane,$x^2+y^2=1$, is an integral curve for the vector field$\mathbf{B}$. I will use this fact to prove that the helicity density must have a zero for any choice of gauge. Let$A$satisfy$dA=B$and suppose$A\wedge B$is non-zero at all points in$D$. Note that$A\wedge B=A(\mathbf{B})\mu_o$, meaning$h=A(\mathbf{B})$is a nowhere vanishing function. Without loss of generality, I will assume$h>0$. Thus, the line integral$I=\oint_C h\frac{dl}{|\mathbf{B}|}$satisfies$I>0$. But, by Stoke's theorem,$I=2\pi\int_0^1g(r)rdr=0$, as is readily verified by directly evaluating the integral. Thus, there can be no such$A\$.

6 minor correction
5 fixed minor mistake